To determine the number of atoms in the products of the reaction between one molecule of iodine (I₂) and two molecules of sodium chloride (NaCl), we first need to identify the products of this reaction.
The reaction can be written as:
\[ \text{I}_2 + 2 \text{NaCl} \rightarrow 2 \text{NaI} + \text{Cl}_2 \]
From this balanced equation, we can identify the products produced: sodium iodide (NaI) and chlorine gas (Cl₂).
Now, let's calculate the number of atoms in the products:
-
Sodium iodide (NaI):
- Each molecule of NaI contains 1 sodium (Na) atom and 1 iodine (I) atom.
- From the balanced equation, 2 molecules of NaI are produced, which gives:
- Sodium: \(2 \times 1 = 2\) Na atoms
- Iodine: \(2 \times 1 = 2\) I atoms
-
Chlorine gas (Cl₂):
- Each molecule of Cl₂ contains 2 chlorine (Cl) atoms.
- From the balanced equation, 1 molecule of Cl₂ is produced, which gives:
- Chlorine: \(1 \times 2 = 2\) Cl atoms
Now, let's sum up all the atoms in the products:
- Sodium atoms: 2
- Iodine atoms: 2
- Chlorine atoms: 2
Total atoms in products = Sodium atoms + Iodine atoms + Chlorine atoms = \(2 + 2 + 2 = 6\) atoms.
Therefore, there are a total of 6 atoms in the products.